Left Bipotent Seminear-Rings

Transcription

1 International Journal of Algebra, Vol. 6, 2012, no. 26, Left Bipotent Seminear-Rings R. Perumal Department of Mathematics Kumaraguru College of Technology Coimbatore, Tamilnadu, India perumalnew R. Balakrishnan Department of Mathematics V.O.Chidambaram College Tuticorin, Tamilnadu, India Abstract In this paper, we introduce the concept of left bipotent seminearrings. We prove certain properties of left bipotent seminear-rings and obtain equivalent conditions for a seminear-ring to be a left bipotent seminear-ring. Mathematics Subject Classification: 16Y60 Keywords: Left normal seminear-ring, Mate function, Idempotent and Nilpotent 1 Introduction Throught this paper, by a seminear-ring we mean a right seminear-ring with an absorbing zero. A seminear-ring is an algebraic system (R, +,.), such that (i) (R, +) is a semigroup (ii) (R,.) is a semigroup (iii)(a + b)c = ac + bc for all a, b, c R. The purpose of the present paper is to introduce the concept of left bipotent seminear-rings and obtain some of their properties. We write ab to denote the product a.b for any two elements a, b in R. For terms and notations used but left undefined in this paper we refer to [ 4] and [ 5]. Notations: We furnish below the notations that we make use of throughout this paper.

2 1290 R. Perumal and R. Balakrishnan 1. E = {e R/e 2 = e} - set of all idempotents of R. 2. L = {x R/x k = 0 for some positive integer k} - set of all nilpotent elements of R. 3. C(R) ={r R/rx = xr for all x R}-centre of R. 2 Preliminary results We freely make use of the following results from [ 4], [ 5] and designate them as K(1), K(2) etc. K(1) A seminear-ring R has no non-zero nilpotent elements if and only if x 2 =0 x = 0 for all x in R. K(2) A map f from R into R is called a mate function for R if x = xf(x)x for all x in R (f(x) is called a mate of x). K(3) A mate function f of R is called a P 3 mate function if for every x in R, xf(x) =f(x)x. K(4) A seminear-ring R is called a P (1, 2) seminear-ring if xr = Rx 2 for all x in R. K(5) A seminear-ring R is called a P k seminear-ring (P k seminear-ring) if there exists a positive integer k such that x k R = xrx (Rx k = xrx) for all x in R. K(6) A seminear-ring R is called left (right) normal a Ra (ar) for each a R. R is normal if it is both left normal and right normal. K(7) Let r be a positive integer. We say that R is a left-r-normal (right-r-normal) seminear-ring if a Ra r (a r R) for each a R. K(8) For any subset A of R we define the radical of A as A = {x R/x k A for some positive integer k}. K(9) If R is a P (1, 2) seminear-ring with a mate function f, then RxRy = Rx Ry = Rxy for all x, y R (Theorem 4.25 of [ 4]). 3 Left Bipotent Seminear-rings In this section, we define the concept of left bipotent seminear-ring and using the above preliminary results, we prove the following main results.

3 Left bipotent seminear-rings 1291 Definition 3.1 We say that a seminear-ring R is left bipotent if Ra = Ra 2 for all a R. Example 3.2 (i) Any Boolean seminear-ring is obviously left bipotent. (ii) Let R = {0,a,b,c,d}. We define the semigroup operations + and. in R as follows. + 0 a b c d 0 0 a b c d a a a b d d b b b b d d c c d d c d d d d d d d. 0 a b c d a 0 a a a a b 0 a b b b c 0 a b b b d 0 a d d d Obviously (R, +,.) is a left bipotent seminear-ring. It is worth noting that this seminear-ring is not Boolean. Theorem 3.3 Homomorphic images of a left normal left bipotent seminearring are also so. Proof: The proof is straight forward. Proposition 3.4 Let R be a left normal left bipotent seminear-ring. Then R has no non-zero nilpotent elements. Proof: Let x R {0}. Since R is left normal, x Rx = Rx 2. Clearly then x = rx 2 for some r in R. Thus x 2 =0 x = 0. Now K(1) guarantees that L = {0}. Hence R has no non-zero nilpotent elements. We furnish below a simple characterization of left bipotent seminearrings. Theorem 3.5 Let R be a left normal seminear-ring. Then R is left bipotent if and only if A = A for every left ideal of R. Proof: For the only if part, let x A. Then there exists some positive integer k such that x k A. Since R is a left normal left bipotent seminear-ring we have, x Rx = Rx 2 x = yx 2 for some y R. Then x = yxx = y(yx 2 )x = y 2 x 3 =... = y k 1 x k RA A. i.e. x A. Therefore A A...(1). But it is obvious that A A...(2). From (1) and (2) we get A = A. For the if part, we observe that Ra 2 is a left ideal of R. If a R, a 3 = aa 2 Ra 2. This implies a Ra 2 = Ra 2. Hence for any x R, xa = x(ya 2 ) Ra 2 for some y R. Therefore Ra Ra 2. Consequently R is left bipotent.

4 1292 R. Perumal and R. Balakrishnan Definition 3.6 (Def in Hanns Joachim Weinert[ 1]) If S is any nonempty subset of R, then the left annihilator of S in R is l(s) ={x R/xs = 0 for all s S}. (We observe that l(s) is a left ideal of R). Proposition 3.7 If R has no non-zero nilpotent elements, l(s) is an ideal for every non empty subset S of R. Proof: It is sufficient to show that l(s)r l(s). Let x l(s) and s S. Then xs = 0 implies (sx) 2 = s(xs)x = 0. Hence sx = 0 by assumption. For any r R, ((xr)s) 2 = xr(sx)rs = 0 implies (xr)s =0. Thusxr l(s). Hence xr l(s). Therefore l(s)r l(s). Proposition 3.8 In a left normal left bipotent seminear-ring R, left annihilators are ideals. Proof: In view of Proposition 3.4, R has no non-zero nilpotent elements. Now Proposition 3.7 guarantees that left annihilators are ideals. Proposition 3.9 Let R be a left normal left bipotent seminear-ring. Then the following are true. (i) x 3 = x 2 implies x 2 = x for every x in R. (ii) l(a) =l(a 2 ) for every a in R. Proof: (i) Suppose x 3 = x 2 for x in R. Since R is left normal left bipotent seminearring, x Rx = Rx 2. Hence x = yx 2 for some y in R, so that x 2 = yx 3 = yx 2 = x. (ii) Let a R, clearly l(a) l(a 2 )...(1). Suppose ya 2 = 0 for some y in R. Now (aya) 2 = ay(aa)ya = aya 2 ya = 0. So, (aya) 3 =(aya) 2 (aya) =0. Thus (aya) 3 =(aya) 2. Hence by (i) aya =(aya) 2 = 0.Now, (ya) 2 = y(aya) = 0 and this implies (ya) 3 =(ya) 2 ya =0. Thus(ya) 3 =(ya) 2. Again by (i) ya =(ya) 2 =0 ya = 0. Hence y l(a) i.e l(a 2 ) l(a)...(2). From (1) and (2) we get l(a) =l(a 2 ). Definition 3.10 An ideal P of R is called a prime ideal if AB P A P or B P holds for any two ideals A, B of R. Definition 3.11 A seminear-ring R is said to be prime, if the {0} is a prime ideal of R. Proposition 3.12 Let R be a left normal left bipotent seminear-ring. If R is prime then R has no non-zero zero divisors.

5 Left bipotent seminear-rings 1293 Proof: Let ab = 0 for a, b in R. Then for any x in R, (bxa) 2 =(bxa)(bxa) = bx(ab)xa = 0. Now Proposition 3.4 demands that bxa = 0 and this implies RbRa = {0}. Since R is prime and Rb, Ra are left ideals we have that, Rb = {0} or Ra = {0}. Since R is left normal, we get either b =0ora =0. Definition 3.13 A seminear-ring R is called left (right) simple if the only non-zero left (right) ideal of R is R itself. Furthermore, R is called simple if it has no non-trivial ideals. Proposition 3.14 Let R be a left bipotent seminear-ring. cancellative then R is simple. If R is right Proof: Let A be a non-zero left ideal of R. If a is any non-zero element of A then Ra = Ra 2.Ifr R then ra = sa 2 for some s in R. Therefore r = sa (as R is right cancellative). Clearly then r A and so A = R. ThusR is simple. Proposition 3.15 Let R be a seminear-ring with a mate function f. Each of the following conditions implies that R is left bipotent. (i) RisaP 1 seminear-ring. (i) f is a P 3 mate function. (ii) Each idempotent in R is central. Proof: (i) Let a R. Since R is a P 1 seminear-ring ar = ara. Since f is a mate function a = af(a)a. Hence af(a) ar = ara. Then there exists y in R such that af(a) =aya. Hence for any r R, ra = r(af(a))a = r(aya)a Ra 2. Thus Ra Ra 2. Obviously Ra 2 Ra. Hence R is left bipotent. (ii) Since f is a P 3 mate function for R, af(a) = f(a)a. Now Ra = Raf(a)a = Rf(a)aa = Rf(a)a 2 Ra 2...(1) Clearly Ra 2 Ra...(2). From (1) and (2) we get Ra = Ra 2 and the result follows. (iii) Since f is a mate function of R. For any a in R, f(a)a is an idempotent, and a = af(a)a = f(a)a 2 (since idempotents are central). Hence Ra = Ra 2 and R is left bipotent. Definition 3.16 A left ideal B of a seminear-ring R is called essential, if B K {0}, for every nonzero left ideal K of R.

6 1294 R. Perumal and R. Balakrishnan Definition 3.17 An element x R is said to be singular, if there exists a nonzero essential left ideal A in R such that Ax = {0}. We denote by S(R) the set of all singular elements of R. Proposition 3.18 Let R be a seminear-ring. Then S(R) ={x R/l(x) is a nonzero essential left ideal in R}. Proof: Let x S(R). Then there exists a nonzero essential left ideal A in R such that Ax = {0}. Therefore, A l(x) and so l(x) is a nonzero essential left ideal in R. Thus S(R) {x R/l(x) is a nonzero essential left ideal in R}. The inclusion the other way is direct since l(x)x = {0}. Proposition 3.19 Let R be a left bipotent seminear-ring. Then L S(R). Proof: Let x L. Then x n = 0 for some n, and since R is left bipotent we have Rx = Rx 2 =... = Rx n = {0}. This gives l(x) =R. Thus l(x) isa nonzero essential left ideal in R. Therefore by Proposition 3.18, x S(R) and the desired result now follows. Proposition 3.20 Let R be a left normal P (1, 2) seminear-ring with a mate function f. Then we have the following: (i) Rx Ry = Rxy for all x, y in R. (ii) R is left bipotent. (iii) R is left-r-normal. Proof: (i) Since R is a P (1, 2) seminear-ring with a mate function f. Now K(9) demands that Rx Ry = Rxy. (ii) For a R. We have Ra = Ra Ra = Raa = Ra 2. That is, Ra = Ra 2 for a R. Hence R is left bipotent. (iii) Since R is a left normal seminear-ring, by (ii) we get R is left bipotent. Therefore a Ra 2 =(Ra)a = Ra 3 =... = Ra r. Hence R is left-rnormal. We conclude our discussion with the following characterisation: Theorem 3.21 Let R be a left normal P(1,1) seminear-ring. Then R is left bipotent if and only if R is P 1 seminear-ring. Proof: Let R be a left bipotent P (1, 1) seminear-ring. Since R is left normal, a Ra = Ra 2 =(Ra)a = ara. Therefore Ra ara. Clearly ara Ra. Thus Ra = ara for every a R. Therefore R is a P 1 seminear-ring. Conversely R be a P 1 seminear-ring. Then Ra = ara = (ar)a = Ra 2. Consequently R is left bipotent.

7 Left bipotent seminear-rings 1295 References [1] Hanns Joachim Weinert, Related representation theorems for rings, semirings, near-rings and seminear-rings by partial transformations and partial endomorbhisms, Proceedings of the Edinburg Mathematical Society, 20, ( ), [2] Jat. J. L and Choudhary. S. C, On Left Bipotent Near-rings, Proceedings of the Edinburg Mathematical Society, 22,(1979), [3] Park. Y. S and Kim. W. J, On Structures of Left Bipotent Near-rings, Kyungbook Math. J, 20(2),(1980), [4] Perumal.R, Balakrishnan.R and Uma.S., Some Special Seminear-ring Structures, Ultra Scientist of Physical Sciences, 23(2)A,(2011), [5] Perumal.R, Balakrishnan.R and Uma.S., Some Special Seminear-ring Structures II, Ultra Scientist of Physical Sciences, 24(1)A,(2012), [6] Pilz Günter, Near Rings North Holland Amsterdam, (1983). [7] Van Hoorn.W.G and Van Rootselaar.B, Fundamental notions in the theory of seminear-rings, compositio mathematic, 18(1967), Received: July, 2012